/****************************************************
 该题充分运用与运算的特点，成对消除，单的最后留下
 ****************************************************/
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
	char str1[10], str2[10];
	int n;
	while(~scanf("%d", &n) )
	{
		getchar();
		gets(str1);
		for(int i=0; i<2*n-2; i++)
		{
			gets(str2);
			for(int j=0; j<7; j++)
			str1[j] = str1[j] ^ str2[j];				//两个字符串各自与运算，可以判定最后剩下的不成对的一个
		}
		printf("%s\n", str1);
	}
}
/*
Description
Alice bought a lot of pairs of socks yesterday. But when she went home, she found that she has lost one of them. Each sock has a name which contains exactly 7 charaters.
Alice wants to know which sock she has lost. Maybe you can help her.
Input
There are multiple cases. The first line containing an integer n (1 <= n <= 1000000) indicates that Alice bought n pairs of socks. For the following 2*n-1 lines, each line is a string with 7 charaters indicating the name of the socks that Alice took back.
Output
The name of the lost sock.
Sample Input
2
aabcdef
bzyxwvu
bzyxwvu
4
aqwerty
eas fgh
aqwerty
easdfgh
easdfgh
aqwerty
aqwerty
2
0x0abcd
0ABCDEF
0x0abcd
Sample Output
aabcdef
eas fgh
0ABCDEF
*/

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